Problem: Is the function given below continuous/differentiable at $x=0$ ? $g(x)=\begin{cases} x^2+5x&,&x<0 \\\\ 2x^2-3x&,&x\geq0 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Solution: Checking for continuity at $x=0$ For the function to be continuous at $x=0$, we need the two-sided limit $\lim_{x\to 0}g(x)$ to exist and be equal to $g(0)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 0^-}g(x)$ and $\lim_{x\to 0^+}g(x)$ exist and are equal to $g(0)$. According to $g$ 's definition, $g(0)=2(0)^2-3(0)=0$. $\lim_{x\to0^-}g(x)$ $x^2+5x$ evaluated at $x=0$ is equal to $0$. Since $x^2+5x$ is continuous, we can be certain that $\lim_{x\to 0^-}g(x)=0$. $\lim_{x\to 0^+}g(x)$ $2x^2-3x$ evaluated at $x=0$ is equal to $0$. Since $2x^2-3x$ is continuous, we can be certain that $\lim_{x\to 0^+}g(x)=0$. We saw that the two one-sided limits exist and are equal to $g(0)$, so the function is continuous at $x=0$. Checking for differentiability at $x=0$ For the function to be differentiable at $x=0$, we need the two-sided limit $\lim_{x\to 0}\dfrac{g(x)-g(0)}{x-0}=\lim_{x\to 0}\dfrac{g(x)-0}{x-0}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 0^-}\dfrac{g(x)-0}{x-0}$ and $\lim_{x\to 0^+}\dfrac{g(x)-0}{x-0}$ exist and have the same value. $\lim_{x\to 0^-}\dfrac{g(x)-0}{x-0}=5$ $\lim_{x\to 0^+}\dfrac{g(x)-0}{x-0}=-3$ The two limits exist, but they are not equal. Therefore, the function is not differentiable at $x=0$. Graphically, the function has a sharp turn at this point. [I would like to see that, please!] In conclusion, the function is continuous but not differentiable at $x=0$.